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Mathematics Question on Differential equations

If dydx=xyx2+y2;y(1)=1;\frac{dy}{dx}=\frac{xy}{x^{2}+y^{2}}; y\left(1\right)=1; then a value of xx satisfying y(x)=ey(x) = e is :

A

3 e\sqrt{3}\,e

B

123 e\frac{1}{2}\sqrt{3}\,e

C

2 e\sqrt{2}\,e

D

e2\frac{e}{\sqrt{2}}

Answer

3 e\sqrt{3}\,e

Explanation

Solution

dydx=xyx2+y2\frac{dy}{dx} = \frac{xy}{x^{2}+y^{2}}
Let y=vxy = vx
dydx=v+x.dvdx\frac{dy}{dx} = v+x. \frac{dv}{dx}
v+xdvdx=xvxx2+v2x2=v1+v2v+x \frac{dv}{dx} = \frac{xvx}{x^{2}+v^{2}x^{2}} = \frac{v}{1+v^{2}}
xdvdx=v1+v2−v=v−v−v31+v2=v31+v2x\frac{dv}{dx}= \frac{v}{1+v^{2}}-v = \frac{v-v-v^{3}}{1+v^{2}} = \frac{v^{3}}{1+v^{2}}
∫v1+v2.dv=∫−dxx\int\frac{v}{1+v^{2}}. dv = \int-\frac{dx}{x}
⇒∫v−3.dv+∫1vdv=−∫dxx\Rightarrow \int v^{-3}.dv+\int \frac{1}{v}dv = -\int \frac{dx}{x}
⇒v−2−2+ℓnv=−ℓnx+λ\Rightarrow \frac{v^{-2}}{-2}+\ell nv = -\ell nx + \lambda
⇒−12v2+ℓn(yx)=ℓnx+λ\Rightarrow - \frac{1}{2v^{2}}+\ell n\left(\frac{y}{x}\right) = \ell nx + \lambda
⇒−12x2y2+ℓny−ℓnx=−ℓnx+λ\Rightarrow - \frac{1}{2} \frac{x^{2}}{y^{2}} +\ell ny - \ell nx = -\ell nx + \lambda
⇒−12+0=λ⇒λ=−12\Rightarrow -\frac{1}{2}+0 = \lambda \Rightarrow \lambda = -\frac{1}{2}
⇒−12x2y2+ℓny+12=0\Rightarrow - \frac{1}{2} \frac{x^{2}}{y^{2}}+\ell ny + \frac{1}{2} = 0 at y=ey = e
⇒−12x2y2+1+12=⇒x22e2=32⇒x2=3e2\Rightarrow - \frac{1}{2} \frac{x^{2}}{y^{2}} +1+\frac{1}{2} = \Rightarrow \frac{x^{2}}{2e^{2}} = \frac{3}{2} \Rightarrow x^{2} = 3e^{2}
∴x=3e\therefore x = \sqrt{3}e